A059480 -id:A059480 - cp's OEIS frontend

A180049 Coefficient triangle of the numerators of the (n-th convergents to) the continued fraction 1/(w+2/(w+3/(w+4/... .

1, 0, 1, 3, 0, 1, 0, 7, 0, 1, 15, 0, 12, 0, 1, 0, 57, 0, 18, 0, 1, 105, 0, 141, 0, 25, 0, 1, 0, 561, 0, 285, 0, 33, 0, 1, 945, 0, 1830, 0, 510, 0, 42, 0, 1, 0, 6555, 0, 4680, 0, 840, 0, 52, 0, 1, 10395, 0, 26685, 0, 10290, 0, 1302, 0, 63, 0, 1, 0, 89055, 0, 82845, 0, 20370, 0, 1926, 0, 75, 0, 1

Offset: 1

Wouter Meeussen, Aug 08 2010

Equivalence to the recurrence formula needs formal proof. This continued fraction converges to 0.525135276160981... for w=1. A conjecture by Ramanujan puts this equal to -1 + 1/(sqrt(e*Pi/2) - Sum_{k>=1} 1/(2k-1)!!). Row sums equal A059480.

			The numerator of 1/(w+2/(w+3/(w+4/(w+5/(w+6/w))))) equals 57w + 18w^3 + w^5.
From _Philippe Deléham_, Nov 06 2013: (Start)
Triangle begins:
      1;
      0,    1;
      3,    0,     1;
      0,    7,     0,    1;
     15,    0,    12,    0,     1;
      0,   57,     0,   18,     0,   1;
    105,    0,   141,    0,    25,   0,    1;
      0,  561,     0,  285,     0,  33,    0,  1;
    945,    0,  1830,    0,   510,   0,   42,  0,  1;
      0, 6555,     0, 4680,     0, 840,    0, 52,  0, 1;
  10395,    0, 26685,    0, 10290,   0, 1302,  0, 63, 0, 1;
  ... (End)
[extended by _M. F. Hasler_, Oct 21 2014]

Hungarian discussion forum

Cf. A059480, A084950, A180047, A180048, A230698.

Table[ CoefficientList[ Numerator[ Together[ Fold[ #2/(w+#1) &, Infinity, Reverse @ Table[ k, {k, 1, n} ] ] ] ], w ], {n, 2, 16} ] or equivalently Clear[ b ]; b[ 0 ]=1; b[ 1 ]=w; b[ n_ ]:=b[ n ] = w b[ n-1 ]+(n+1) b[ n-2 ]; Table[ CoefficientList[ b[ k ]//Expand, w ], {k, 0, 14} ]

t=x-w;for(n=1,12,t=substpol(t,x,w+n/x);print(Vecrev(numerator(substpol(t,x,w))))) \\ M. F. Hasler, Oct 21 2014

b(0)=1; b(1)=w; b(n) = w*b(n-1) + (n+1)*b(n-2) (conjecture).

A070895 Triangle read by rows where T(n+1,k)=T(n,k)+n*T(n-1,k) starting with T(n,n)=1 and T(n,k)=0 if n

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 10, 6, 4, 1, 1, 26, 18, 8, 5, 1, 1, 76, 48, 28, 10, 6, 1, 1, 232, 156, 76, 40, 12, 7, 1, 1, 764, 492, 272, 110, 54, 14, 8, 1, 1, 2620, 1740, 880, 430, 150, 70, 16, 9, 1, 1, 9496, 6168, 3328, 1420, 636, 196, 88, 18, 10, 1, 1, 35696, 23568

Offset: 0

Henry Bottomley, May 23 2002

For n>k+1, T(n,k) is a multiple of k+2.

Eigentriangle of inverse of (-1)^(n-k)*A094587. Row sums are A187044. - Paul Barry, Mar 02 2011

			Rows start: 1; 1,1; 2,1,1; 4,3,1,1; 10,6,4,1,1; etc.
Triangle begins
1,
1, 1,
2, 1, 1,
4, 3, 1, 1,
10, 6, 4, 1, 1,
26, 18, 8, 5, 1, 1,
76, 48, 28, 10, 6, 1, 1,
232, 156, 76, 40, 12, 7, 1, 1
Production matrix begins
1, 1,
1, 0, 1,
1, 1, 0, 1,
2, 1, 1, 0, 1,
4, 3, 1, 1, 0, 1,
10, 6, 4, 1, 1, 0, 1,
26, 18, 8, 5, 1, 1, 0, 1,
76, 48, 28, 10, 6, 1, 1, 0, 1,
232, 156, 76, 40, 12, 7, 1, 1, 0, 1
Inverse begins
1,
-1, 1,
-1, -1, 1,
0, -2, -1, 1,
0, 0, -3, -1, 1,
0, 0, 0, -4, -1, 1,
0, 0, 0, 0, -5, -1, 1,
0, 0, 0, 0, 0, -6, -1, 1
- _Paul Barry_, Mar 02 2011

Columns include A000085, A000932, A059480. Right hand columns effectively include A000012 (twice), A000027, A005843, A028552. Cf. A062323 for a triangle with similar formulas.

T(n, k+1)=(T(n, k-1)-T(n-1, k))/k for 0

A247376 Triangular array: row n gives the coefficients of the polynomial p(n,x) defined in Comments.

Original entry on oeis.org

1, 2, 2, 3, 5, 5, 15, 8, 8, 35, 33, 13, 80, 131, 48, 21, 171, 409, 279, 34, 355, 1180, 1375, 384, 55, 715, 3128, 5335, 2895, 89, 1410, 7858, 18510, 17029, 3840, 144, 2730, 18851, 58253, 78609, 35685, 233, 5208, 43629, 171059, 317758, 243873, 46080, 377, 9810

Offset: 0

Clark Kimberling, Oct 23 2014

The polynomial p(n,x) is the numerator of the rational function given by f(n,x) = 1 + (2*x + 1)/f(n-1,x), where f(0,x) = 1.
(Sum of numbers in row n) = A059480(n+1) for n >= 0.
(Column 1) is essentially A000045 (Fibonacci numbers).

Clark Kimberling, Rows 0..100, flattened

Cf. A059480, A000045.

z = 15; f[x_, n_] := 1 + (2 x + 1)/f[x, n - 1]; f[x_, 1] = 1;
t = Table[Factor[f[x, n]], {n, 1, z}]
u = Numerator[t]
TableForm[Table[CoefficientList[u[[n]], x], {n, 1, z}]] (* A247376 array *)
Flatten[CoefficientList[u, x]] (* A247376 sequence *)

rown(n) = if (n==0, 1, 1 + (2*x+1)/rown(n-1));
tabl(nn) = for (n=0, nn, print(Vecrev(numerator(rown(n))))); \\ Michel Marcus, Oct 28 2014

f(0,x) = 1/1, so that p(0,x) = 1
f(1,x) = (2 + 2 x)/1, so that p(1,x) = 2 + 2 x;
f(2,x) = (3 + 5 x)/(2 + 2 x), so that p(2,x) = 3 + 5 x.
First 6 rows of the triangle of coefficients:
1
2    2
3    5
5    15    8
8    35    33
13   80    131   48

A371943 Number of permutations that end with a consecutive pattern 123, and avoid consecutive patterns 123 and 213 elsewhere.

Original entry on oeis.org

0, 0, 0, 1, 2, 10, 28, 116, 388, 1588, 5960, 25168, 102856, 453608, 1985008, 9163360, 42486128, 205065136, 1000056928, 5035366208, 25689681760, 134588839648, 715328668736, 3889568161408, 21463055829568, 120839175460160, 690344333849728, 4015753752384256

Offset: 0

Yixin Lin, Apr 13 2024

nonn

(This can be proved by observing the possible positions of n.)

			For n=0, 1, 2, there are no permutations ending with 123. Hence, a(0)=a(1)=a(2)=0. For n=3, a(3)=1 since 123 is the only permutation that ends with 123. For n=4, a(4)=2 with qualified permutations 3124, 4123. For n=5, a(5)=10 with qualified permutations 14235, 15234, 24135, 25134, 34125, 35124, 43125, 45123, 53124, 54123.

Sergi Elizalde and Yixin Lin, Penney's game for permutations, arXiv:2404.06585 [math.CO], 2024.

Cf. A059480.

a:= proc(n) option remember; `if`(n<3, 0, `if`(n=3, 1,
      2*a(n-1)+2*(n-2)*(a(n-2)-a(n-3))-(n-2)*(n-3)*a(n-4)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Apr 13 2024

a[n_]:=a[n]=If[n<3, 0, If[n==3, 1, 2*a[n-1]+2*(n-2)*(a[n-2]-a[n-3])-(n-2)*(n-3)*a[n-4]]]; Table[a[n], {n,0,27}] (* Stefano Spezia, Apr 13 2024 *)
RecurrenceTable[{a[n] == 2*a[n-1] + 2*(n-2)*(a[n-2] - a[n-3]) - (n-2)*(n-3)*a[n-4], a[0] == 0, a[1] == 0, a[2] == 0, a[3] == 1}, a, {n, 0, 30}] (* Vaclav Kotesovec, Apr 14 2024 *)
nmax = 30; FullSimplify[CoefficientList[Series[-1 + E^(x*(2 + x)/2) * (1 + x) + E^((1 + x)^2/2) * Sqrt[Pi/2] * (2 + x)*(Erf[1/Sqrt[2]] - Erf[(1 + x)/Sqrt[2]]), {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Apr 14 2024 *)

def aList(len):
    b = [0, 0, 0, 1, 4]
    a = [0, 0, 0, 1, 2]
    for i in range(4, len):
        b.append(b[i] + i * b[i - 1])
        a.append(a[i] + i * a[i - 1] + b[i])
    return a
print(aList(27))

a(0)=a(1)=a(2)=0, a(3)=1, a(4)=2; a(n) = a(n-1)+(n-1)*a(n-2)+b(n-1), where b(n) = b(n-1)+(n-1)*b(n-2) is the same sequence as A059480, up to the first initial terms. Here, our b(n) has initial terms 0, 0, 0, 1, 4.
From Vaclav Kotesovec, Apr 14 2024: (Start)
a(n) ~ c * n^((n+1)/2) * exp(sqrt(n) - n/2), where c = exp(-1/4) / sqrt(2) - exp(1/4) * sqrt(Pi) * erfc(1/sqrt(2)) / 2 = 0.189615662815288097469466802437...
E.g.f.: -1 + exp(x*(2 + x)/2) * (1 + x) + exp((1 + x)^2/2) * sqrt(Pi/2) * (2 + x) * (erf(1/sqrt(2)) - erf((1 + x)/sqrt(2))). (End)

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A180049 Coefficient triangle of the numerators of the (n-th convergents to) the continued fraction 1/(w+2/(w+3/(w+4/... .

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A070895 Triangle read by rows where T(n+1,k)=T(n,k)+n*T(n-1,k) starting with T(n,n)=1 and T(n,k)=0 if n

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A247376 Triangular array: row n gives the coefficients of the polynomial p(n,x) defined in Comments.

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A371943 Number of permutations that end with a consecutive pattern 123, and avoid consecutive patterns 123 and 213 elsewhere.

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